KVL in an RLC Circuit

Apply KVL rule similarly ,

Voltage across L is L di/dt

Voltage across C is 1/C Integral (i) dt

Instead of solving algebraic equations, now you have to solve differential equations. Now the challenge come, you got to do a bit of study here, no short cut here! Laplace tried to find a simpler method , he managed to make the lives of electrical engineering students a lot more miserable !

Thanks Bravo88. Solving differential equations - not a problem :-)

Could we write the set of equations using the 's' variable of the Laplace equation? So the voltage drop for the For the inductor it would be sL*i and for the capacitor (1/sC)*i. Am I correct?

Yes, if you know Laplace, how come you don't know L di/dt and 1/C integral (i)dt?

I have solved Laplace equations on a purely mathematical basis earlier, I have studied these equations too, but they never stick to my mind because I have never practically solved RLC circuits using KVL.

Looks like a homework problem... But I would convert everything into the S domain using Laplace transforms. In the S domain Z(s) = Ls for inductors, 1/(Cs) for capacitors, and resistors don't change. Once here you can solve like regular circuit then do the inverse Laplace to get back to the time domain.

But to get to the equations Bravo has lead you so directly to, you simply need to solve the voltage across each of the components (see image) to apply KVL. You don't need to use the sign designations that I put because we do not know if they are correct! So long as you keep them consistent within all three loops you will be fine. It gets tedious but all EE's have to do it. Example of one equation for L1 (Loop1, sorry for the confusion I probably caused with L=Loop and L=Inductor... I'm sure you'll figure it out haha) would be 0 = -(Vcos(at+b))+ (1/C1 integral (i) dt) + (i*R1)

Thanks. :-)